 # 2006 Chapter Solutions

```2006
MATHCOUNTS CHAPTER
SPRINT ROUND
1. We have two identical blue boxes
and three identical red boxes. The
two blue boxes together weigh the
same as the three red boxes. The
red boxes each weigh 10 ounces.
Therefore, the three red boxes
weigh 10 × 3 = 30 ounces. The two
blue boxes weigh the same, i.e., 30
ounces. Therefore, one blue box
weighs
30
= 15. Ans.
2
2. Three black chips have been placed
on the game board as shown.
We are asked to find the number of
the square where a fourth chip
should be placed so that it shares
neither a row nor a column with any
of the existing chips. Rows 1, 3 and
4 have chips. Therefore, one must
be in Row 2. Columns 1, 3, and 4
have chips. Therefore, one must be
in Column 2. The number in Row 2,
Column 2, is 5. Ans.
3. We are asked to find the greatest
number of cubes, with edge length 1
inch, which can be placed into a
rectangular box measuring 3 inches
by 3 inches by 9 inches. Clearly, it
is not desirable to have any spaces.
We can fill the bottom of the
rectangle with 3 rows of 9 cubes
each. We can also have a total of 3
of these sets of 3 rows of 9 cubes.
So what we are computing here is
simply the volume.
V = lwh = 9 × 3 × 3 = 27 × 3 = 81
Ans.
4. 6 ×
1 6
=
Ans.
7 7
5. Krista put 1 cent into her new bank
on Sunday. On Monday she put 2
cents in and on Tuesday she put 4
cents in. Each day she doubles the
amount that she put in the day
before. We must find the day of the
week that the total amount of money
in her bank first exceeded \$2. Let’s
look at the first couple of days…
DAY AMOUNT
TOTAL
1
1
1
2
2
3
3
4
7
4
8
15
Day 1 has a total of 21 – 1 = 2 – 1 =
1 cent.
Day 2 has a total of 22 – 1 = 4 – 1 =
3 cents.
Day 3 has a total of 23 – 1 = 8 – 1 =
7 cents.
Day 4 has a total of 24 – 1 = 16 – 1
= 15 cents.
There is a pattern here. (It’s a good
thing to recognize powers of 2 and
one less than each power of 2!)
Now we only need to find the first
power of 2 that is greater than 200
to find the number of days. Starting
with the power 1, we have 2, 4, 8,
16, 32, 64, 128, 256. So
28 – 1 = 256 – 1 = 255 and it takes 8
days, starting with Sunday, for us to
get over 200 cents. Since Sunday
is the first day, it must also be the
eighth day. Sunday Ans.
6. We are given a table of positions,
how many people have these
positions and how much they are
paid.
TITLE
#
SALARY
Pres.
1
\$130,000
Vice-Pres.
5
\$ 90,000
Director
10
\$ 75,000
Assoc. Dir.
6
\$ 50,000
\$ 23,000
To find the median salary value, first
determine the number of
employees.
1 + 5 + 10 + 6 + 37 = 59
The median is the middle one or
number 30 (i.e., 29 + 29 + 1 = 59)
specialists, and that is the lowest
salary, the median is within that
number, and the salary is \$23,000.
Ans.
7. Each side of Square ABCD is
doubled in length to form Square
EFGH.
D
C
having AB blood.
15 3
= = 75% Ans.
20 4
10. A type of cat food recommends that
1
a cat have a daily serving of
3
ounce of dry cat food per pound of
body weight. A particular cat is fed
3
2
ounces of dry food following the
3
recommendations. To find its
weight, just determine how many
thirds are in 3
The perimeter of square EFGH is 40
cm. This makes each side of the
square 10 cm. Since square EFGH
had each side doubled in length
from square ABCD, a side of square
ABCD is
1
of 10 or 5. The area of
2
square ABCD is 5 × 5 = 25. Ans.
8. The time zones of New York and
Denver are different by 2 hours. A
train leaves New York at 2 p.m.
(New York time) and arrives in
Denver 45 hours later. This
equivalent to saying that the train
leaves New York at noon, Denver
time, and arrives in Denver 45 hours
later. If it were to arrive in Denver
48 hours later (instead of 45) it
would be exactly two days later or
noon in Denver time. But since it
takes only 45 hours instead of 48, it
must arrive three hours earlier, or 9
a.m. Ans.
9. We are given a chart of male and
female patients and how many have
each type of blood. Since we are
asked to find what percent of
patients with type AB blood are
male, we need only consider the
column dealing with type AB blood.
That column shows 15 males having
type AB blood and 5 females having
AB blood for a total of 20 people
2
2 11
. 3 =
3
3 3
Thus, the cat must weigh 11
pounds. 11 Ans.
11. Roger has exactly one of each of
the first 22 states’ new U.S.
quarters. The quarters were
released in the same order that the
states joined the union and the
graph shows the number of states
that joined the union in each
decade. The graph shows that 12
states joined the union in the
12 6
=
of Roger’s quarters are
22 11
from states that joined the union in
6
Ans.
11
12. We are given the sequence:
0, 1, 1, 3, 6, 9, 27, …
0 is the first term and each
subsequent term is produced by
each successive integer beginning
with 1. Thus 1 = 0 + 1.
1=1×1
3=1+2
6=3×2
9=6+3
27 = 9 × 3
What is the value of the first term
that is greater than 125? This
shouldn’t take long…just continue…
31 = 27 + 4
124 = 31 × 4 (gotta go one more!!!)
129 = 124 + 5
129 Ans.
13. Carolyn, Julie and Roberta share
\$77 in a ratio of 4:2:1, respectively.
We are asked to determine how
x = the amount of money Roberta
4x + 2x + 1x = 77
7x = 77
x = 11
Carolyn had 4 times as much as
Roberta. 4 × 11 = 44 Ans.
For the second column, certainly red
may not be one of the choices.
Choose yellow (light grey in black &
white) on top and green (medium
grey in black & white) on the bottom.
Then all the rest of the colors are
pre-determined. Look at the
following drawing:
14. The weight A is balanced by the four
weights, 9, 3, 3, and 1. B is
balanced by the 5 weights 9, 9, 3, 3,
and 1. We are asked to find the
minimum number of weights it would
take to balance the total weight of A
+ B where the available weights are
1, 1, 3, 3, 9, 9, 27 and 27. The
weight of A is 9 + 3 + 3 + 1 = 16.
The weight of B is 9 + 9 + 3 + 3 + 1
= 25. Thus, the weight of A + B =
16 + 25 = 41. To find the smallest
number of weights used as the
equivalent, start by using the most
weights of 27. 41 – 27 = 14
We can use one weight of 27. We
can also use 1 weight of 9 leaving
14 – 9 or 5 to go. 5 can be handled
by 1 weight of 3 and 2 weights of 1
for a total of 1 + 1 + 1 + 2 = 5
weights. Can we do any better?
Replacing a larger weight with
smaller weights (even if we had
weights to the total.
5 Ans.
You see that green has colored the
top hexagon in the third column.
Suppose we’d chosen red. Then
green would have to be either the
second or third hexagon and the
yellow the other. In any event, we’d
have green touching green which is
illegal. Now suppose we’d chosen
yellow for the top hexagon. Then
we have choices of red and green
for the second and third hexagons
and again, we’d violate the rules
because we’d still have green
touching green. Similarly, for the
rest of the hexagons we have a predetermined set of colors. In the
same way, if we go back and
choose green on top and yellow on
bottom for the second column of
hexagons, as in the following
picture,
15. a = 2, b = 3, c = 4
(b – c)2 + a(b + c) =
(3 – 4)2 + 2(3 + 4) =
(-1)2 + 2(7) = 1 + 14 = 15 Ans.
16. The first red hexagon (very dark
grey in black & white) touches the
next two hexagons as in the
following picture.
all the rest of the hexagons are predetermined. 2 Ans.
17. One quart of paint is exactly enough
for two coats of paint on a 9-foot by
10-foot wall. This means that one
quart of paint covers 9 × 10 × 2 =
180 square feet. A 10-foot by 12foot wall is 10 × 12 = 120 square
feet.
120 12 2
=
=
Ans.
180 18 3
18. Ten unit cubes are glued together
as shown.
We are asked to find the surface
area. Since these are unit cubes,
there is an area of 1 for each cube
face. Start by counting up the
number of cube faces in the front.
These are shown in red in the next
figure.
Similarly, the bottom also shows 4
cube faces.
10 + 10 + 4 + 4 + 3 + 3 = 34 Ans.
19. A graph shows the total distance
Sam drove from 6 a.m. to 11 a.m.
Though the graph shows the
following information:
6 a.m. – 0 miles
7 a.m. – 40 miles
8 a.m. – 60 miles
9.a.m. – 100 miles
10 a.m. – 120 miles
11 a.m. – 160 miles, the most
important thing to see is that
Sam drove 160 miles in 5 hours.
His average speed is
160
= 32 Ans.
5
20. A stock loses 10% of its value on
There are 3 + 4 + 3 = 10 cube faces
in the front. Similarly, there are 10
cube faces in the back. Now count
up the number of cube faces visible
from the right as shown in blue in
the figure below.
There are 3. Similarly, there are 3
on the left. Now count up how many
are shown from the top. These are
shown in green in the next figure.
There are 3 on top and then
fragments to the left and right of the
top row. But these two must add up
to 1 cube face since the top row’s 3
cube faces are sitting on 4 cube
faces of the second row. 3 + 1 = 4.
Monday. On Tuesday it loses 20%
of the value it had at the end of the
day on Monday. Let x = the cost of
the stock on Monday morning. At
the end of Monday, it has lost 10%
of its price or 0.1x.
x – 0.1x = 0.9x
On Tuesday it lost 20% of the price
at the end of Monday.
0.2 × 0.9x = 0.18x
0.9x – 0.18x = 0.72x
Thus, the stock now costs 72% of
what it did on Monday morning and
has lost 100% - 72% = 28% of its
value. 28 Ans.
21. The numbers 1, 2, and 3 are written
in nine unit squares according to the
following rules:
-- Each of the numbers appears
three times and only one number is
placed in each of the nine unit
square.
-- Each number is in a unit square
a unit square with the same number.
-- The sum of the numbers in the
leftmost column and the sum of the
numbers in the top row are each 7.
24. Splitting 2004 into 200 and 4 yields
The second rule states that a
number must be next to (horizontally
or vertically) the same number and
the third rule states that the sum of
the numbers in the leftmost column
and in the top row are each 7.
7=3+3+1
7=3+2+2
Place the first one in the first column
and the second one in the first row
and everything falls out.
two integers with a common factor
greater than 1. The same holds for
2005 and 2006. We are asked to
find the first odd-numbered year
after 2006 that has this property.
Consider 200 and 7. 7 is not a
divisor of 200.
Consider 200 and 9. 9 = 3 × 3 and
3 is not a divisor of 200. If we move
on to 2011 we have 201 and 1.
Clearly this doesn’t satisfy, but
looking at 201 says that 3 must be a
factor (2 + 0 + 1 = 3 and 3 is
divisible by 3) so 2013 will satisfy
our requirements. 2013 Ans.
25. Two consecutive even numbers are
3 + 2 + 1 + 1 = 7 Ans.
22. Five balls are numbered 1-5 and
placed in a bowl. Josh chooses a
ball, looks at it and then puts it back
in the bowl. Then he chooses a ball
again. We are asked to determine
the probability that the product of
the two numbers will be even and
greater than 10. There are 5 × 5 =
25 combinations. Of these, the
even products greater than 10 are:
3 × 4 = 12
4 × 3 = 12
4 × 4 = 16
4 × 5 = 20
5 × 4 = 20
or 5 different combinations (actually
only 3 combinations but 2 of them
can be switched around).
5 1
=
Ans.
25 5
23. When a certain negative number is
multiplied by six, the result is the
same as 20 less than the original
number.
Let x be the original number. Then:
6x = x – 20
5x = -20
x = -4 Ans.
each squared. The difference of the
squares is 60.
Let x = the first even number.
Let x+2 = the second even number.
Then (x + 2)2 – x2 = 60
x2 + 4x + 4 – x2 = 60
4x = 56
x = 14
14 + 14 + 2 = 30
**Note that they didn’t specify which
square was subtracted from the
other square so we have to
try it the other way.
x2 – (x + 2)2 = 60
x2 – (x2 +4x +4) = 60
x2 – x2 -4x -4 = 60
-4x -4 = 60
-4x = 64
4x = -64
x = -16
-16 + -16 + 2 = -30
Thus, both 30 and -30 are answers.
**30; –30; or 30 and –30 were
accepted. Ans.
26. We are asked to find the area of the
pentagon shown here with sides of
length 15, 20, 27, 24, and 20 units.
First draw a horizontal line (in red)
creating the rectangle of dimensions
20 × 24. The area of this rectangle
is:
20 × 24 = 480
Next, draw a line from the top of
vertical line on the left hand side to
the vertex where the lines of length
27 and 20 meet (the blue line). This
gives us a right triangle with two
sides of 15 and 20. Since this is
obviously a multiple of a 3, 4, 5 right
triangle, the hypotenuse is 25. The
area of the 15, 20, 25 right triangle
is:
½ × 15 × 20 = 150
This leaves one more right triangle
whose hypotenuse is 25 and one of
whose sides is 7. If you don’t know
about 7, 24, 25 right triangles, just
work it out.
72 + x2 = 252
49 + x2 = 625
x2 = 625 – 49 = 576
x = 24
The area of the 7, 24, 25 right
triangle is:
½ × 7 × 24 = 84
Totaling everything up:
480 + 150 + 84 =
630 + 84 = 714 Ans.
27. I start my bike ride at 20 miles per
hour and later continue at only 12
miles per hour. I travel a total of
122 miles in 8 hours. How long did I
feel good (i.e., travel at 20 miles per
hour)?
Let x = the number of hours I
traveled at 20 miles per hour.
Then 8-x is the number of hours I
traveled at 12 miles per hour.
20x + 12(8 – x ) = 122
20x + 96 – 12x = 122
8x = 26
x=
26 13
=
Ans.
8
4
28. Charlie made a list of the page
number of the last page he finished
numbers after 8 days came up to
432 pages. Charlie read the same
amount of pages each day.
Let x be the number of pages that
x + 2x + 3x + 4x + 5x + 6x + 7x + 8x
= 432
36x = 432
x = 12
So Charlie read 12 pages per day.
12 × 8 = 96 Ans.
29. We are given that each distinct letter
in the equation MATH = COU + NTS
is replaced by a different digit
chosen from 1 through 9 in such a
way that the resulting equation is
true.
COU
+NT S
MATH
H = 4 and we are asked to find the value
of the greater of C and N.
C + N (+ a possible carry) gives us a
value of A ones and M tens. M can only
be 1 since the maximum value of two
digits and a carry is 19.
COU
+NTS
1AT4
Now look at O + T (+ a possible carry) =
T If O were 0 (zero) then O + T = T but
O must be some digit between 1 and 9.
Therefore we must have a carry.
O+1+T=T
There will have to be a carry here for
this to make sense.
O + 1 + T = T + 10
O=9
C9U
+NTS
1AT4
Now let’s deal with U and S.
U + S = 14
14 = 9 + 5
14 = 8 + 6
14 = 7 + 7
9 + 5 doesn’t work since O is already 9
and each letter corresponds to a unique
digit. 7 + 7 doesn’t work for the same
reason.
So we know that U = 8 and S = 6 or vice
versa; it doesn’t really matter.
C98
+NT6
1AT4
So, what numbers are left? Just 2, 3, 5,
and 7. Clearly C + N + 1 > 10
We have the sum of two numbers plus 1
greater than 10. It could be 3 + 7 + 1 or
5 + 7 + 1. In any case the greatest
value of C or N must be 7. Ans.
30. A gear turns 33
1
times in a
3
minute. Another gear turns 45 times
mark on each pointing north. We
are asked to determine how many
seconds will it take before the two
gears next have both their marks
pointing due north. The first gear
has its mark face north every
60
60
3
180
=
= 60 ×
=
1 100
100 100 =
33
3
3
9
seconds.
5
The second gear has its mark face
north every
60 4
=
seconds.
45 3
Put these two values into the same
denominator.
9
=
5
4
=
3
27
15
20
15
We need to find the least common
multiple of 27 and 20.
Neither number has any factor in
common.
27 = 3 × 3 × 3
20 = 2 × 2 × 5
Therefore, the LCM is 27 × 20 = 540
540
= 36 Ans.
15
TARGET ROUND
48
= 12 Ans.
4
2. The rules for traveling the maze
require us to only follow a path
where the score is between 2 and
14. We have the following
possibilities but must check after
each operation that we stay within
the limit.
7 + -2 + 3 + -4 + 5
7 + -2 – 3 – -4 + 5
7 + -2 – 3 + -4 – 5
7 – -2 + 3 – -4 + 5
7 – -2 + 3 + -4 – 5
7 – -2 – 3 – -4 – 5
Starting with the first possibility:
7 + -2 = 5; 5 + 3 = 8; 8 – 4 = 4;
4+5=9
Second possibility:
7 + -2 = 5; 5 – 3 = 2 STOP; we’re
not between 2 and 14.
Third possibility:
7 + -2 = 5; 5 – 3 = 2; 2 + -4 = -2
STOP; we’ve gone under.
Fourth possibility:
7 – -2 = 9; 9 + 3 = 12; 12 – -4 = 16;
STOP; we’ve gone over.
Fifth possibility:
7 – -2 = 9; 9 + 3 = 12; 12 + -4 = 8;
8–5=3
Sixth possibility:
7 – -2 = 9; 9 – 3 = 6; 6 – -4 = 10;
10 – 5 = 5
The three possibilities are 9, 3, and
5. We are asked to find the lowest
score and that is 3. Ans.
3. In the figure below, side AE of
rectangle ABDE is parallel to the xaxis and side BD contains the point
C. The vertices of the triangle ACE
are given in the figure.
1. Brass contains 80% copper and
20% zinc. Henri’s brass trumpet
contains 48 ounces of copper. We
need to determine how much zinc.
The ratio is 80% to 20% or 4 to 1.
Therefore, for every 4 ounces of
copper there must be 1 ounce of
zinc.
We are asked to find the ratio of the
area of triangle ACE to the area of
rectangle ABDE. The base of
triangle ACE is 4 – 1 = 3. The
height of triangle ACE is 3 – 1 = 2.
The area of triangle ACE is:
½×3×2=3
The width of rectangle ABDE is the
length of side AE which, as before,
is 3. The height of rectangle ABDE
is the same as the height of triangle
ACE, or 2. The area of rectangle
ABDE is 3 × 2 = 6.
3 1
=
Ans.
6 2
(But then, any time you have as two
the third point is on the side of the
rectangle opposite the base of the
triangle, the area of the triangle will
always be
1
of the area of the
2
rectangle!)
4. The dimensions of A4 paper are
0.21 meters by 0.297 meters. (As
an aside, I’ve actually used this
paper!) The area of one sheet of A4
paper is 0.21 × 0.297 = 0.06237
square meters. There are 21 sheets
so the total area is 21 × 0.06237 =
1.30977 ≈ 1.3 Ans.
5. The arithmetic mean of A, B, and C
is 10. This means A + B + C = 30.
A=B–6
C=B+3
A + C = 2B – 3
A + B + C = A + C + B = 30 =
(2B – 3) + B
3B – 3 = 30
3B = 33
B = 11
C = B + 3 = 11 + 3 = 14 Ans.
6. The values 1 through 26 are
assigned to A through Z,
respectively. A 9-digit code is
created for each letter using prime
factorization. The first digit of a
letter’s code is the number of times
2 is used as a factor; the second
digit is the number of times 3 is
used as a factor and so on. We are
given 6 9-digit codes and asked to
determine what word this set of
codes spells. The first 9-digit code
is:
001000000
The only 1 is in the 3rd column
where the 3rd prime, 5, is used as
the factor so 5 is the first value or E.
The second 9-digit code is:
000000100
The only 1 is in the 7th column
where the 7th prime, 17, is used as
the factor so 17 is the second value
or Q.
The third 9-digit code is:
010100000
The 1’s are the second and fourth
prime or 3 and 7.
3 × 7 = 21 or U
The fourth 9-digit code is:
000000000
The only number with no prime
factors is 1 or A.
The fifth 9-digit code is:
210000000
22 × 3 = 4 × 3 = 12 or L
The sixth 9-digit code is:
000000010
The only 1 is in the 8th column
where the 8th prime, 19, is used as
the factor so 19 is value or S
EQUALS Ans.
7. A quiz has 15 easy questions and
15 hard questions. Easy questions
are worth 4 points each and hard
questions are worth 10 points each.
Sam earns 92 points on the quiz.
We are asked to find the greatest
number of hard questions that he
First ask what is the largest multiple
of 10 that is less than 92? Certainly
this is 9. But that means he’d get 90
points from the hard ones leaving 2
points for the easy ones. Since an
easy question is 4 points this won’t
work. Try 8 hard ones. Then there
are 92 – 80 = 12 points or 3 easy
questions. 8 Ans.
8. Eli throws five darts at a circular
target as shown and each one lands
within one of the four regions.
TEAM ROUND
1. In 1992 1200 lire was the same as
\$1.50. Therefore 1 lire was the
same as:
The point values of the red (outer),
blue, green and yellow (inner) areas
are 1, 2, 4, and 6 points,
respectively. We are asked to find
the least score greater than five
points that is not possible when the
point values of the five darts are
What values can we make with only
5 darts?
6=1+1+1+1+2
7=1+1+1+2+2
8=1+1+2+2+2
9=1+2+2+2+2
10 = 2 + 2 + 2 + 2 + 2
11 = 4 + 1 + 2 + 2 + 2
12 = 4 + 2 + 2 + 2 + 2
13 = 4 + 4 + 1 + 2 + 2
14 = 4 + 4 + 2 + 2 + 2
15 = 4 + 4 + 4 + 1 + 2
16 = 4 + 4 + 4 + 2 + 2
17 = 4 + 4 + 4 + 4 + 1
18 = 4 + 4 + 4 + 4 + 2
19 = 4 + 4 + 4 + 1 + 6
20 = 4 + 4 + 4 + 2 + 6
21 = 4 + 4 + 1 + 6 + 6
22 = 4 + 4 + 2 + 6 + 6
23 = 4 + 1 + 6 + 6 + 6
24 = 4 + 2 + 6 + 6 + 6
25 = 1 + 6 + 6 + 6 + 6
26 = 2 + 6 + 6 + 6 + 6
27 = ?
27 is the first one that we can’t
make. Note that there is a limit to
the checking. We can never get to
a point value of 31.
Note: A pattern emerges
making the counting easier. Every 1
can change to a 2 to create the next
number; then every 2,2 can change
to a 4,1 for the next number; every
4,1 can change to a 4,2 for the next
number, every 4,2 can change to a
6,1 for the next number; and every
6,1 can change to a 6, 2 for the next
number. Seeing this pattern makes
the listing of the possibilities easier.
27 Ans.
1200 1
=
1.5
x
1200x = 1.5
x=
1.5
= 0.00125
1200
1 lire is the same as \$0.00125.
1,000,000 lire is the same as
\$0.00125 × 1,000,000 = \$1250 Ans.
2
of the
3
height from which it falls. We
are asked to determine after how
many bounces the ball first rises
less than 30 cm if it is dropped
from a height of 243 cm. So
consider dropping it from a
height of 243 cm. After the first
bounce it comes back up to:
2
243 ×
= 162 cm
3
After the second bounce it comes
back up to:
2
162 ×
= 108 cm
3
After the third bounce it comes
back up to:
2
108 ×
= 72 cm
3
After the fourth bounce it comes
back up to:
2
72 ×
= 48 cm
3
After the fifth bounce it comes
back up to:
2
48 ×
= 32 cm
3
And, after the sixth bounce it
comes back up to:
2. A ball bounces back up
2
1
= 21 cm
3
3
So it’s on the sixth bounce that
the ball first rises less than 30
cm. 6 Ans.
32 ×
3. In 2003, the Dodonpa roller
coaster had a maximum speed of
106.9 miles per hour. We are
asked to look at the graph and
determine how many roller
that was faster than the Dodonpa
roller coaster. Looking at the
graph we see the Dodonpa roller
coaster at a height of about 175
feet. There is only one more
point to the right at about 120+
miles per hour at around 400
feet. 1 Ans.
4. Marika purchased her house with
a loan for 80% of the price and
paid the remaining \$49,400 with
her savings. This \$49,400 was
100% - 80% = 20% of the
1
purchase price (or )
5
49400 × 5 = 247,000 Ans.
5. One interior angle of a convex
polygon is 160 degrees. The rest
of the interior angles of the
polygon are each 112 degrees.
The number of degrees in a
polygon of n sides or n angles is:
180 × (n – 2)
160 + (112 × (n – 1)) = 180 × (n – 2)
160 + 112n – 112 = 180n – 360
180n – 360 = 112n + 48
180n – 112n = 360 + 48
68n = 408
n=
408
= 6 Ans.
68
6. A 25 passenger bus rents for \$110.
A 40 passenger bus rents for \$170.
We are asked to find the minimum
cost for renting enough buses for a
school trip with 475 passengers.
A 25 passenger bus costs:
110
= \$4.40 per person
25
a 40 passenger bus costs:
170 17
=
= \$4.25 per person
40
4
The bigger bus is cheaper per
person. Try using as many of those
as possible.
475
= 11 big buses + 35 people
40
We’d need 12 buses but we’d have
some empty seats.
12 × 170 = \$2040
What about using ten 40 passenger
buses and three 25 passenger
buses so we don’t have any empty
seats…
(10 × 170) + (3 × 110) =
1700 + 330 = \$2030
There’s no way we can do any
better because we’ll end up using
more of the more expensive buses.
2030 Ans.
7. Arpan and Tomika want to place
straw in the flower beds surrounding
their house. Every angle is a right
represent the flower beds.
The width of the flower beds is 2
feet. Each bale of straw costs \$2.75
and covers 9 square feet of ground.
We are asked to calculate how
much it will cost to cover the flower
beds with straw if only whole bales
of straw may be purchased.
First, we have to figure out how
many square feet of flower bed we
have. The top part of the flower bed
is 62 feet in width but you also have
to add on the 2 feet in width on
either side. Thus, the area of the
flower bed shaded in red above is
(62 + 2 + 2) × 2 = 66 × 2 = 132
Next is the left vertical part of the
bed. How long is that? Well,
looking at the right side (by the
yellow colored flower bed), you can
see that this part is 20 feet. But also
by the green part you can see that
there is an additional 6 feet of
vertical length so the left side is 20 +
6 = 26 feet but you also have to add
an additional 2 feet for the flower
bed edge. Thus, the area of the
blue part is:
(26 + 2) × 2 = 28 × 2 = 56
Next is the right vertical part. This is
easier. Only the top is extra flower
bed but that was accounted for in
the red section. Therefore the area
of the yellow section is just:
20 × 2 = 40
Finally how much area is in the
green portion? Well, we know it’s
two feet wide but what about the
that the top side is 62 feet long. The
bottom is just like that, but there are
20 feet missing on the right and 4
feet missing near the left.
62 – (20 + 4) = 62 – 24 = 38
Thus, while we don’t know how
much of the 38 feet is in the left part
colored green or in the right part, it
doesn’t matter. This is equivalent to
a rectangle of 38 × 2, so the total
area is 76 square feet.
The total square footage is:
132 + 56 + 40 + 76 = 304
are asked to find the total cost of the
three meals. Here are the menus:
Entrée:
Hot dog: \$1.25
Hamburger: \$1.65
Chicken: \$1.80
Pizza: \$2.25
304
= 33.77777777777777777
9
Finally the third most expensive
meal.
If we go with chicken, it would be 45
cents cheaper. If we go with coffee,
it would be 30 cents cheaper and if
we go with cookie it would be 65
cents cheaper. So maybe the
coffee? Well, that would make this
meal a total of 20 + 30 = 50 cents
cheaper than the most expensive
meal. But if we switch in chicken for
pizza and go back up to the juice
the meal would only be 45 cents
cheaper so the third most expensive
meal is chicken, juice and pudding.
1.80 + 1.25 + 1.30 – 0.20 = 4.15
Thus, the total value for all three
meals is:
4.60 + 4.40 + 4.15 = 13.15 Ans.
bales of straw. Since we must buy
whole bales we will need 34 bales.
The straw costs 34 × \$2.75 =
\$93.50
With 6% tax this is \$93.50 × 1.06 =
\$99.11 Ans.
8. A “value Meal” consists of one
below. The value meal price is
calculated by summing up the prices
of the three individual items and
subtracting 20 cents from the total.
The customer orders exactly one of
each of the three most expensive
value meal combinations and we
Drink:
Coffee: \$0.75
Soda: \$1.05
Juice: \$1.25
Dessert:
Pudding: \$1.30
To find the most expensive meal,
let’s take the most expensive item
from each category.
That would be pizza, juice and
pudding for a total of:
2.25 + 1.25 + 1.30 – 0.20 (for the
“value” ) = 4.60
What’s the next most expensive
meal? Well, if we go from pizza to
chicken it would be 45 cents
cheaper. If we go from juice to
soda, it would be 20 cents cheaper.
If we go from pudding to cookie, it
would be 65 cents cheaper. Go with
the soda, so the second most
expensive meal is pizza, soda and
pudding.
2.25 + 1.05 + 1.30 – 0.20 = 4.40
9. A 6 by 6 grid of 36 unit squares is
completely covered in T-shaped
pieces consisting of four unit
squares. The pieces can not
overlap but may extend over the
side of the 6 by 6 grid. We are
asked to find the fewest number of
pieces required to cover the grid.
Since these pieces are T-shaped
the best we can do is to try and
interweave them as much as
possible and leave the number of
portions extending outside the grid
as the absolute minimum. The
following grid illustrates how to do
that.
The way to see this (without trying
to do the drawing first) is as follows:
The 6 by 6 grid has 36 unit squares
in it. Each of the T-shaped pieces
occupies 4 squares.
36
= 9 pieces but we know we’re
4
going to extend over the edge so
the best we can shoot for is 10 and
we’ve proven we can do that. 10
Ans.
10. Three friends have a full bag of jelly
1
beans. Mike took
of the jelly
3
1
beans in the full bag. Zac took
2
of the jelly beans in the full bag.
Kary took what was left. Mike ate
1
1
of his jelly beans, Zac ate
of
2
3
his jellybeans and Kary ate all of
hers. Mike and Zac now have a
total of 45 jellybeans together and
we re asked to find out how many
the jelly beans and then ate
1
of
3
1
of
2
1
of the
6
entire original bag of jelly beans left.
1
1
of the bag and ate
of
2
3
them leaving
1  1 1 1 1 1
−  ×  = − = of the
2  3 2 2 6 3
original bag of jelly beans left.
Between the two of them they now
have 45 jellybeans which must be
1
of the bag of jellybeans. Thus,
2
the total number of jellybeans that
were originally in the bag was:
45 × 2 = 90 jellybeans. Between
Mike and Zac they originally had
1 1 5
+ =
of the bag leaving just
3 2 6
1
of the bag for Kary.
6
1
× 90 = 15 Ans.
6
```